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Which of the following function is surjective but not injective
$f : R \to R$ $f (x) = x^4 + 2x^3 - x^2 + 1$
$f : R \to R$ $f (x) = x^3 + x + 1$
$f : R \to R^+ f (x) =$ $\sqrt {1 + {x^2}} \,$
$f : R \to R f (x) = x^3 + 2x^2 - x + 1$
Solution
$f (x) = x^4 + 2x^3 – x^2 + 1 \to A$ polynomial of degree even will always be into say
$f (x) = a_0x^{2n} + a_1 x^{2n-1} + a_2 x^{2n – 2} + …. + a_{2n}$
$\mathop {Limit}\limits_{x \to \, \pm \,\infty } \,f (x) $
$=\mathop {Limit}\limits_{x \to \, \pm \,\infty } \,$ $[x^{2n} $ $\left( {{a_0}\, + \,\frac{{{a_1}}}{x}\, + \,\frac{{{a_2}}}{{{x^2}}}\, + \,….\, + \frac{{{a_{2n}}}}{{{x^{2n}}}}} \right)$
$= \left[ \begin{gathered} \,\infty \,\,\,\,\,if\,{a_0} > 0 \hfill \\ – \,\infty \,\,if\,{a_0}\, < 0 \hfill \\ \end{gathered} \right.$
Hence it will never approach $\infty / – \infty$
$f (x) = x^3 + x + 1 \Rightarrow f ‘(x) = 3x^2 + 1$ – injective as well as surjective
$f (x) =$$\sqrt {1 + {x^2}} \,$ – neither injective nor surjective
$f (x) = x^3 + 2x^2 – x + 1\Rightarrow f ‘(x) =3x^2 + 4x – 1 \Rightarrow D > 0$
Hence $f (x)$ is surjective but not injective.